What is the solution for y^'-(x^2y-y)/(y+4)=0,y(3)=1 ?

The solution for y^'-(x^2y-y)/(y+4)=0,y(3)=1 is y+4ln(y)=(x^3)/3-x-5

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y^'-(x^2y-y)/(y+4)=0,y(3)=1

y^{^{'}} - \frac{x ^{2} y - y}{y + 4} = 0, y (3) = 1

y + 4 ln (y) = \frac{x ^{3}}{3} - x - 5

Solution steps

y^{'} (x) - \frac{x ^{2} y - y}{y + 4} = 0

Solve separable ODE:

y + 4 ln (y) = \frac{x ^{3}}{3} - x - 5

y + 4 ln (y) = \frac{x ^{3}}{3} - x - 5

(e^{m} + 1)^{2} e^{- m} d t + (e^{t} + 1)^{3} e^{- t} d m = 0

\frac{d y}{d x} + cos (x) y - cos (x) = 0

\frac{d y}{d x} = \frac{1 + x}{x y ^{5}}

2 x^{2} y^{^{'}} = y^{2}

y^{^{'}} - e^{x} y = 0

What is the solution for y^'-(x^2y-y)/(y+4)=0,y(3)=1 ?
The solution for y^'-(x^2y-y)/(y+4)=0,y(3)=1 is y+4ln(y)=(x^3)/3-x-5